📚 Mastering Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebra and graphs to measurement and problem-solving. This revision guide explains how to recognise a quadratic equation, solve it by factorisation, completing the square, and the quadratic formula, and how to use the discriminant to decide the nature of the roots.
二次方程贯穿 IGCSE 数学课程的各个部分，从代数、图像到测量与应用题。本复习指南将讲解如何识别二次方程、如何用因式分解法、配方法和求根公式求解，以及如何利用判别式判断根的性质。
1. What Is a Quadratic Equation? | 什么是二次方程？
be written in the general form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是未知数最高次数为 2 的多项式方程。它总能写成一般形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。
bx + c is a parabola, and solving the equation means finding the x-values where the parabola crosses the x-axis.
如果 a = 0，方程就变成了一次方程，因此 a ≠ 0 是二次方程成立的关键条件。二次函数 y = ax² + bx + c 的图像是一条抛物线，解方程就是求抛物线与 x 轴交点的 x 值。
- x² + 5x + 6 = 0
- 2x² − 3x − 2 = 0
- x² − 9 = 0
2. Standard Form and Key Vocabulary | 标准形式与关键术语
term ax² is called the quadratic term, bx is the linear term, and c is the constant term. The coefficient a controls the shape and direction of the parabola, while c gives the y-intercept.
在标准形式 ax² + bx + c = 0 中，ax² 称为二次项，bx 称为一次项，c 称为常数项。系数 a 决定抛物线的形状和开口方向，常数项 c 表示图像与 y 轴的交点。
ic; this often makes factorisation simpler. When a is not 1, you need extra care in factorising or using the quadratic formula.
当 a = 1 时，二次式称为首一多项式，这通常会使因式分解更简单。当 a 不等于 1 时，因式分解或使用求根公式时需要更加仔细。
ax² + bx + c = 0, a ≠ 0
3. Solving by Factorisation: The Zero Product Principle | 因式分解法：零乘积原理
Factorisation relies on the zero product principle: if the product of two factors is zero, then at least one factor must be zero. Therefore, once we write ax² + bx + c as (px + q)(rx + s), we can set each bracket equal to zero separately.
因式分解法依据零乘积原理：如果两个因式的乘积为零，那么至少有一个因式为零。因此，一旦我们把 ax² + bx + c 写成 (px + q)(rx + s) 的形式，就可以分别令每个括号等于零。
For a monic quadratic x² + bx + c, look for two numbers that multiply to c and add to b. For example, to factorise x² + 5x + 6, we need two numbers that multiply to 6 and add to 5.
对于首一二次式 x² + bx + c，要寻找两个数，使它们的乘积等于 c，和等于 b。例如，要对 x² + 5x + 6 进行因式分解，我们需要两个数，乘积为 6，和为 5。
When a ≠ 1, use the ac method or trial and error. The factorisation method is fastest when the quadratic has simple integer roots, but it does not work neatly for every equation.
当 a ≠ 1 时，可使用 ac 法或尝试法。当二次方程有简单的整数根时，因式分解法最快，但它并非对每个方程都简便。
4. Worked Example: Factorisation | 例题：因式分解
s that multiply to +6 and add to +5. These are +2 and +3, so x² + 5x + 6 = (x + 2)(x + 3).
解方程 x² + 5x + 6 = 0。我们需要找到两个数，乘积为 +6，和为 +5。这两个数是 +2 和 +3，因此 x² + 5x + 6 = (x + 2)(x + 3)。
Set each factor to zero: x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
令每个因式为零：x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。
x = −2 or x = −3
and +1. Split the middle term: 2x² − 4x + x − 2 = 2x(x − 2) + 1(x − 2) = (x − 2)(2x + 1).
对于 2x² − 3x − 2，找出两个数，乘积为 2 × (−2) = −4，和为 −3：−4 和 +1。拆中间项：2x² − 4x + x − 2 = 2x(x − 2) + 1(x − 2) = (x − 2)(2x + 1)。
Set each bracket equal to zero: x − 2 = 0 or 2x + 1 = 0, so x = 2 or x = −1/2.
令每个括号等于零：x − 2 = 0 或 2x + 1 = 0，所以 x = 2 或 x = −1/2。
x = 2 or x = −1/2
5. Completing the Square: Turning x² + bx into a Perfect Square | 配方法：将 x² + bx 化为完全平方
p)². For x² + bx, we add (b/2)² to create a perfect square: x² + bx + (b/2)² = (x + b/2)².
配方法利用恒等式 x² + 2px + p² = (x + p)²。对于 x² + bx，我们加上 (b/2)² 来构造完全平方：x² + bx + (b/2)² = (x + b/2)²。
This method is powerful because it works for every quadratic and also reveals the turning point of the parabola. If the coefficient of x² is not 1, first divide or factor out a so that the coefficient of x² becomes 1.
这种方法非常强大，因为它适用于所有二次方程，并且还能揭示抛物线的顶点。如果 x² 的系数不是 1，需要先除以或提取系数 a，使 x² 的系数变为 1。
x² + bx + (b/2)² = (x + b/2)²
h, k). Completing the square is therefore especially useful for maximum and minimum problems.
y = (x + h)² + k 的顶点为 (−h, k)。因此，配方法在最大值和最小值问题中特别有用。
6. Worked Example: Completing the Square | 例题：配方法
he constant: x² + 6x = −2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 7, so (x + 3)² = 7.
用配方法解方程 x² + 6x + 2 = 0。移常数项：x² + 6x = −2。两边加上 (6/2)² = 9：x² + 6x + 9 = 7，因此 (x + 3)² = 7。
Take square roots: x + 3 = ±√7, so x = −3 ± √7. These two roots are exact values; no rounding is needed unless the question asks for decimal answers.
开平方：x + 3 = ±√7，所以 x = −3 ± √7。这两个根是精确值；除非题目要求小数答案，否则不需要取近似值。
x = −3 ± √7
. Then x² + 4x = −3/2. Add (4/2)² = 4: x² + 4x + 4 = 5/2, so (x + 2)² = 5/2. Thus x = −2 ± √5/2.
对于
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图